## Deep Learning From Scratch IV: Gradient Descent and Backpropagation

This is part 4 of a series of tutorials, in which we develop the mathematical and algorithmic underpinnings of deep neural networks from scratch and implement our own neural network library in Python, mimicing the TensorFlow API. Start with the first part: I: Computational Graphs.

Generally, if we want to find the minimum of a function, we set the derivative to zero and solve for the parameters. It turns out, however, that it is impossible to obtain a closed-form solution for $W$ and $b$. Instead, we iteratively search for a minimum using a method called gradient descent.

As a visual analogy, imagine yourself standing on a mountain and trying to find the way down. At every step, you walk into the steepest direction, since this direction is the most promising to lead you towards the bottom.

If taking steep steps seems a little dangerous to you, imagine that you are a mountain goat (which are amazing rock climbers).

Gradient descent operates in a similar way when trying to find the minimum of a function: It starts at a random location in parameter space and then iteratively reduces the error $J$ until it reaches a local minimum. At each step of the iteration, it determines the direction of steepest descent and takes a step along that direction. This process is depicted for the 1-dimensional case in the following image.

As you might remember, the direction of steepest ascent of a function at a certain point is given by the gradient at that point. Therefore, the direction of steepest descent is given by the negative of the gradient. So now we have a rough idea how to minimize $J$:

1. Start with random values for $W$ and $b$
2. Compute the gradients of $J$ with respect to $W$ and $b$
3. Take a small step along the direction of the negative gradient
4. Go back to 2

Let’s implement an operation that minimizes the value of a node using gradient descent. We require the user to specify the magnitude of the step along the gradient as a parameter called learning_rate.

The following image depicts an example iteration of gradient descent. We start out with a random separating line (marked as 1), take a step, arrive at a slightly better line (marked as 2), take another step, and another step, and so on until we arrive at a good separating line.

# Backpropagation

In our implementation of gradient descent, we have used a function compute_gradient(loss) that computes the gradient of a $loss$ operation in our computational graph with respect to the output of every other node $n$ (i.e. the direction of change for $n$ along which the loss increases the most). We now need to figure out how to compute gradients.

Consider the following computational graph:

By the chain rule, we have

$$\frac{\partial e}{\partial a} = \frac{\partial e}{\partial b} \cdot \frac{\partial b}{\partial a} = \frac{\partial e}{\partial c} \cdot \frac{\partial c}{\partial b} \cdot \frac{\partial b}{\partial a} = \frac{\partial e}{\partial d} \cdot \frac{\partial d}{\partial c} \cdot \frac{\partial c}{\partial b} \cdot \frac{\partial b}{\partial a}$$

As we can see, in order to compute the gradient of $e$ with respect to $a$, we can start at $e$ an go backwards towards $a$, computing the gradient of every node’s output with respect to its input along the way until we reach $a$. Then, we multiply them all together.

Now consider the following scenario:

In this case, $a$ contributes to $e$ along two paths: The path $a$, $b$, $d$, $e$ and the path $a$, $c$, $d$, $e$. Hence, the total derivative of $e$ with respect to $a$ is given by:

$$\frac{\partial e}{\partial a} = \frac{\partial e}{\partial d} \cdot \frac{\partial d}{\partial a} = \frac{\partial e}{\partial d} \cdot \left( \frac{\partial d}{\partial b} \cdot \frac{\partial b}{\partial a} + \frac{\partial d}{\partial c} \cdot \frac{\partial c}{\partial a} \right) = \frac{\partial e}{\partial d} \cdot \frac{\partial d}{\partial b} \cdot \frac{\partial b}{\partial a} + \frac{\partial e}{\partial d} \cdot \frac{\partial d}{\partial c} \cdot \frac{\partial c}{\partial a}$$

This gives us an intuition for the general algorithm that computes the gradient of the loss with respect to another node: We perform a backwards breadth-first search starting from the loss node. At each node $n$ that we visit, we compute the gradient of the loss with respect do $n$’s output by doing the following for each of $n$’s consumers $c$:

• retrieve the gradient $G$ of the loss with respect to the output of $c$
• multiply $G$ by the gradient of $c$’s output with respect to $n$’s output

And then we sum those gradients over all consumers.

As a prerequisite to implementing backpropagation, we need to specify a function for each operation that computes the gradients with respect to the inputs of that operation, given the gradients with respect to the output. Let’s define a decorator @RegisterGradient(operation_name) for this purpose:

Now assume that our _gradient_registry dictionary is already filled with gradient computation functions for all of our operations. We can now implement backpropagation:

For each of our operations, we now need to define a function that turns a gradient of the loss with respect to the operation’s output into a list of gradients of the loss with respect to each of the operation’s inputs. Computing a gradient with respect to a matrix can be somewhat tedious. Therefore, the details have been omitted and I just present the results. You may skip this section and still understand the overall picture.

If you want to comprehend how to arrive at the results, the general approach is as follows:

• Find the partial derivative of each output value with respect to each input value (this can be a tensor of a rank greater than 2, i.e. neither scalar nor vector nor matrix, involving a lot of summations)
• Compute the gradient of the loss with respect to the node’s inputs given a gradient with respect to the node’s output by applying the chain rule. This is now a tensor of the same shape as the input tensor, so if the input is a matrix, the result is also a matrix
• Rewrite this result as a sequence of matrix operations in order to compute it efficiently. This step can be somewhat tricky.

### Gradient for negative

Given a gradient $G$ with respect to $-x$, the gradient with respect to $x$ is given by $-G$.

### Gradient for log

Given a gradient $G$ with respect to $log(x)$, the gradient with respect to $x$ is given by $\frac{G}{x}$.

### Gradient for sigmoid

Given a gradient $G$ with respect to $\sigma(a)$, the gradient with respect to $a$ is given by $G \cdot \sigma(a) \cdot \sigma(1-a)$.

### Gradient for multiply

Given a gradient $G$ with respect to $A \odot B$, the gradient with respect to $A$ is given by $G \odot B$ and the gradient with respect to $B$ is given by $G \odot A$.

### Gradient for matmul

Given a gradient $G$ with respect to $AB$, the gradient with respect to $A$ is given by $GB^T$ and the gradient with respect to $B$ is given by $A^TG$.

### Gradient for add

Given a gradient $G$ with respect to $a + b$, the gradient with respect to $a$ is given by $G$ and the gradient with respect to $b$ is also given by $G$, provided that $a$ and $b$ are of the same shape. If $a$ and $b$ are of different shapes, e.g. one matrix $a$ with 100 rows and one row vector $b$, we assume that $b$ is added to each row of $a$. In this case, the gradient computation is a little more involved, but I will not spell out the details here.

### Gradient for reduce_sum

Given a gradient $G$ with respect to the output of reduce_sum, the gradient with respect to the input $A$ is given by repeating $G$ along the specified axis.

## Example

Let’s now test our implementation to determine the optimal weights for our perceptron.

Notice that we started out with a rather high loss and incrementally reduced it. Let’s plot the final line to check that it is a good separator:

If you have any questions, feel free to leave a comment. Otherwise, continue with the next part: V: Multi-Layer Perceptrons

By | 2017-11-18T13:09:59+00:00 August 26th, 2017|Artificial Intelligence, Deep Learning, Machine Learning, Python, TensorFlow|13 Comments
• FunnyBretzel

Good job explaining this, Daniel!

• thecity2

This is cool.

• Ayush Joshi

Would really like to see what exactly is being calculated when you say gradient of J w.r.t W and b. Like you have the chain rule diagram showing the nodes a,b,c,d,e, I think if you could actually write the nodes J,W,X,B appropriate to the graph and show what exact gradient is calculated that would make this way more understandable for me. Thanks for the amazing post !

• Daniel Sabinasz

Good point. The reason that I use letters is the fact that the algorithm is independent of the computational graph that we use to feed it, so it could be used to optimize any function J with respect to any input nodes, regardless of how they contribute to J. But you’re right that it would be helpful to apply it to the examples that we encountered already. I grant that the explanation of backpropagation is a little short. I’ll probably work on an update in the future.

For now, imagine every input matrix to every node. For every individual input matrix, we compute a matrix of the same shape as that input matrix. The entry at (i, j) in that matrix tells us how J changes as a function of the entry (i, j) in the input matrix, that is, dJ / dM(i, j).

Take a look at the final perceptron graph from lesson 2. Here, we would first compute dJ / dSigma, then dJ/(XW+b), then dJ/db and dJ/dXW, then dJ/dW and dJ/dX. The shapes of these derivatives correspond to the shape of what’s in the denominator.

I hope this helps.

• Ayush Joshi

Hi Daniel,

Your explanation helped and as I mentioned earlier I think would understand the derivative process more clearly when i saw the math for each step. As you mention that it gets tedious, can you recommend a source for where one can find the relevant steps?

• Ayush Joshi

I think it also brings to light the advancements in symbolic algebra that made neural networks popular again.

• Daniel Sabinasz

First of all, it is important to realize that backpropagation is a meta-algorithm that is independent of the individual operations and how their gradients are computed. It only presupposes that there is a method for each operation that turns a gradient with respect to a node’s output into gradients with respect to the node’s inputs.

Thus, you should first try to understand backpropagation as a meta-algorithm on general computational graphs without considering any neural network stuff whatsoever.

Once you are at that point, you can read the section “Gradient of each operation” that defines these functions. True, I haven’t developed the math for each operation. Maybe start with the gradient computation for “log” and “multiply”. These are the easier ones. Once you understand how they come about, you should be able to work out the math for the other gradients yourself. It’s a nice exercise. And I’ll admit that I don’t know them anymore, I worked them out once on paper and decided that the computations are too tedious to include in the blog.

Understanding the principles is, however, more important than the individual computations. You only derive them once, code them and then you’ll probably never think about them again in your entire life.

• shambolix

Thanks for this tutorial, it’s pitched at just the right level and is exactly what I was looking for 🙂 I did find this section a bit difficult to follow though, because I was initially confused why the backwards propagation for a given node was looking at the derivative of the consumer rather than at its inputs. I found it quite helpful to focus on one step first and keeping the loss in mind e.g. dl/dd = dl/de * de/dd thus when stepping from e to d, and knowing dl/de already, you need to evaluate de/dd which is the consumer operation from the point of view of d. Once I got my head around that, the rest became clear.

One question though – intuitively I would have expected the gradient function to have been a method of the Operation so that it is defined and accessed in the same way as the mathematical operation itself. Is there a particular reason for why this code (and presumably tensorflow) uses the registry and decorator to bind them instead?

• Daniel Sabinasz

The reason I did it this way is because TensorFlow does it this way. I do not know what their reason is, actually. I would have preferred to put it into the operation class.

• Sanford Chung

why is that it is impossible to obtain a closed-form solution for WW and bb using the closed form. don’t we have the derivative chain figured out , then set it to zero?

• Daniel

Yes, you can write it down and set it to zero, but it is impossible to solve this equation for the weights.

• Mahmoud S. Ahmed

When I follow the tutorial’s code, in the example part I get this error.

Step: 0 Loss: 65.7448384796

—————————————————————————
KeyError Traceback (most recent call last)
in ()
35 if step % 10 == 0:
36 print(“Step:”, step, ” Loss:”, J_value)
—> 37 session.run(minimization_op, feed_dict)
38
39 # Print final result

in run(self, operation, feed_dict)
13 else:
14 node.inputs = [input_node.output for input_node in node.input_nodes]
—> 15 node.output = node.compute(*node.inputs)
16
17 if type(node.output) == list:

in compute(self)
11 def compute(self):